calculus - Proof that $Γ'(1) = -γ$?

I know that $Γ'(1) = -γ$, but how does one prove this?
Starting from the basics, we have that:

$$Γ(x) = \int_0^\infty e^{-t} t^{x-1} dt$$

How do we differentiate this? How do we then find that

$$Γ'(1) = \int_0^\infty e^{-t} \log(t) dt$$

and how would one solve this integral?

Answer

Taken from this answer:

By the recursive relation $\Gamma(x+1)=x\Gamma(x)$, we get

$$\small{\log(\Gamma(x))=\log(\Gamma(n+x))-\log(x)-\log(x+1)-\log(x+2)-\dots-\log(x+n-1)}\tag{1}$$
Differentiating $(1)$ with respect to $x$, evaluating at $x=1$, and letting $n\to\infty$ yields
$$\begin{align} \frac{\Gamma'(1)}{\Gamma(1)}&=\log(n)+O\left(\frac1n\right)-\frac11-\frac12-\frac13-\dots-\frac1n\\ &\to-\gamma\tag{2} \end{align}$$
A discussion of why $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))=\log(x)+O\left(\frac1x\right)$, and a proof of the log-convexity of $\Gamma(x)$, is given in the answer cited above. It is easy to accept since $\log(\Gamma(n+1))=\log(\Gamma(n))+\log(n)$.

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Another Approach

$$\begin{align} \int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t &=\lim_{n\to\infty}\int_0^n\log(t)\,\left(1-\frac{t}{n}\right)^n\,\mathrm{d}t\tag{3a}\\ &=\lim_{n\to\infty}n\int_0^1(\log(t)+\log(n))\,(1-t)^n\,\mathrm{d}t\tag{3b}\\ &=\lim_{n\to\infty}\left(\frac{n}{n+1}\log(n)+n\int_0^1\log(1-t)\,t^n\,\mathrm{d}t\right)\tag{3c}\\ &=\lim_{n\to\infty}\left(\frac{n}{n+1}\log(n)-\frac{n}{n+1}H_{n+1}\right)\tag{3d}\\[6pt] &=-\gamma\tag{3e} \end{align}$$
Explanation:
$\text{(3a)}$: Monotone Convergence and $e^{-t}=\lim\limits_{n\to\infty}\left(1-\frac tn\right)^n$
$\text{(3b)}$: substitute $t\mapsto nt$
$\text{(3c)}$: substitute $t\mapsto1-t$
$\text{(3d)}$: integrate by parts $\int_0^1\log(1-t)\,t^n\,\mathrm{d}t=-\frac1{n+1}\int_0^1\frac{1-t^{n+1}}{1-t}\,\mathrm{d}t$
$\text{(3e)}$: $\gamma=\lim\limits_{n\to\infty}H_n-\log(n)$